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Chapter 18: Problem 103

It took \(2.30\) min using a current of \(2.00 \mathrm{~A}\) to plate out all the silver from \(0.250 \mathrm{~L}\) of a solution containing \(\mathrm{Ag}^{+}\). What was the original concentration of \(\mathrm{Ag}^{+}\) in the solution?

Short Answer

Expert verified
The original concentration of \(Ag^+\) ions in the solution was \(0.0114\,\mathrm{M}\).

Step by step solution

01

Determine the charge transferred using the current and time

Given that the current is \(I = 2.00\,\mathrm{A}\), and the time duration is \(t = 2.30\,\mathrm{min}\), we can find the total charge (Q) transferred during the plating process. Convert the time to seconds and use the formula \(Q = It\), which gives: \(Q = (2.00\,\mathrm{A})(2.30\,\mathrm{min} \times 60\,\mathrm{s/min})\) \(Q = 276\,\mathrm{C}\)
02

Calculate the moles of silver ions plated out

Use Faraday's law, which relates the quantity of a substance undergoing electrolysis to the charge transferred: \(n = \frac{Q}{zF}\) Here, n represents the moles of silver ions plated out, z is the charge number (+1 for \(Ag^+\) ions), and F is the Faraday constant (\(96485\,\mathrm{C/mol}\)). Calculate the moles of silver ions: \(n=\frac{276\,\mathrm{C}}{1\times96485\,\mathrm{C/mol}}\) \(n = 0.00286\,\mathrm{mol}\)
03

Find the original concentration of Ag(+) in the solution

We are given the volume of the solution \(V = 0.250\,\mathrm{L}\), and we have calculated the moles of \(Ag^+\) ions plated out (n). Divide the moles (n) by the volume (V) to find the initial concentration of \(Ag^+\) ions in the solution: \(c_{Ag^+} = \frac{n}{V}\) \(c_{Ag^+} = \frac{0.00286\,\mathrm{mol}}{0.250\,\mathrm{L}}\) \(c_{Ag^+} = 0.0114\,\mathrm{M}\) So, the original concentration of \(Ag^+\) ions in the solution was \(0.0114\,\mathrm{M}\).

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Most popular questions from this chapter

A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ} &=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ} &=0.000 \mathrm{~V} \end{aligned} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartmentcontains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, HA, at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

The overall reaction and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is \(\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \quad \mathscr{E}^{\circ}=1.10 \mathrm{~V}\) For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)
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